Proving naturality in both arguments

Question

I am having a hard time showing that two functors are adjoint. Consider as an example \begin{align} t&: \mathbf{Ab} \longrightarrow \mathbf{T},\\ i&: \mathbf{T} \longrightarrow \mathbf{AB}. \end{align} Where $\mathbf{T}$ denotes the full subcategory of $\mathbf{Ab}$ consisting of all torsion abelian groups. $t$ takes any abelian group, $A$ to its largest torsion subgroup, $t(A)$, and any morphism to its restriction on $t(A)$. I have already shown that this is a well defined functor. $i$ denotes the inclusion of $\mathbf{T}$ into $\mathbf{Ab}$.

Now I want to show that $t$ is right adjoint to $i$. From the definition I know this would mean to construct a natural isomorphism $$\mathrm{Hom}_\mathbf{Ab}(i(M), N) \xrightarrow{\cong} \mathrm{Hom}_\mathbf{T}(M, t(N)).$$ For $M$ a torsion abelian group and $N$ an abelian group.

Since $\mathbf{T}$ is full in $\mathbf{Ab}$ we have $\mathrm{Hom}_\mathbf{T}(M, t(N)) = \mathrm{Hom}_\mathbf{Ab}(M, t(N))$. Moreover $i(M) = M$ in $\mathbf{Ab}$. Hence I must construct a natural isomorphism $$\mathrm{Hom}_\mathbf{Ab}(M, N) \xrightarrow{\cong} \mathrm{Hom}_\mathbf{Ab}(M, t(N)).$$

I have shown that any map from $M \rightarrow N$ must factor through $M \rightarrow t(N)$, i.e., there is an isomorphism from $\mathrm{Hom}_\mathbf{Ab}(M, N)$ to $\mathrm{Hom}_\mathbf{Ab}(M, t(N))$ by composition with the inclusion $t(N) \hookrightarrow N$. Then for a morphism $f: A \rightarrow M$ the below square commutes:

$$ \begin{array}{ccc} \mathrm{Hom}_\mathbf{Ab}(M, t(N))& \longrightarrow & \mathrm{Hom}_\mathbf{Ab}(M, N)\\ \downarrow& & \downarrow \\ \mathrm{Hom}_\mathbf{Ab}(A, t(N))& \longrightarrow & \mathrm{Hom}_\mathbf{Ab}(A, N)\\ \end{array} $$

That is composing with the inclusion of $t(N)$ into $N$ yields a natural isomorphism at least in the first argument by my understanding. How should I go about proving naturality in the last argument? Also I do think I am struggling with bifunctors in general, so If someone have an intuitive way to think about them, please let me know. Thanks :)

Answer 1

More generally, let $\mathcal{D}$ be a full subcategory of $\mathcal{C}$ with the property that it is coreflective: for every $X \in \mathcal{C}$ there shall be some $T(X) \in \mathcal{D}$ with a morphism $\iota_X : T(X) \to X$, such that for every morphism $Y \to X$, where $Y \in \mathcal{D}$ there is a unique morphism $Y \to T(X)$ such that the evident triangle commutes (universal property). Think of this situation as a (vast) generalization of the torsion subgroup example.

The universal property exactly states $$\alpha_{Y,X} : \hom_{\mathcal{D}}(Y,T(X)) \to \hom_{\mathcal{C}}(Y,X),\quad f \mapsto \iota_X \circ f$$ is an isomorphism (bijection) for all $X \in \mathcal{C}$ and $Y \in \mathcal{D}$.

If $g : X \to X'$ is a morphism in $\mathcal{C}$, we define $T(g) : T(X) \to T(X')$, using the universal property, by the equation $\iota_{X'} \circ T(g) = g \circ \iota_X$. Then it follows easily that $T$ is indeed a functor.

Let us check that $\alpha$ is natural in $X$, i.e. that for all $g : X \to X'$ the diagram $$\require{AMScd} \begin{CD} \hom_{\mathcal{D}}(Y,T(X)) @>{\alpha_{Y,X}}>> \hom_{\mathcal{C}}(Y,X) \\ @V{T(g)_*} VV @VV{g_*}V \\ \hom_{\mathcal{D}}(Y,T(X'))@>>{\alpha_{Y,X'}}> \hom_{\mathcal{C}}(Y,X') \end{CD} $$ commutes. This is really just a matter of using the definitions. For $f : Y \to T(X)$ we compute $$\begin{align*} g_*(\alpha_{Y,X}(f)) & = g \circ \iota_X \circ f \\ &= \iota_{X'} \circ T(g) \circ f \\ &= \iota_{X'} \circ T(g)_*(f) \\ & = \alpha_{Y,X'}(T(g)_*(f)) \end{align*}$$

And to check naturality in $Y$ (which is even more easy), let us take any morphism $h : Y' \to Y$ and verify that $$\require{AMScd} \begin{CD} \hom_{\mathcal{D}}(Y,T(X)) @>{\alpha_{Y,X}}>> \hom_{\mathcal{C}}(Y,X) \\ @V{h^*} VV @VV{h^*}V \\ \hom_{\mathcal{D}}(Y',T(X))@>>{\alpha_{Y',X}}> \hom_{\mathcal{C}}(Y',X) \end{CD} $$

commutes. Well, for $f : Y \to T(X)$ we compute $$h^*(\alpha_{Y,X}(f)) = (\iota_X \circ f) \circ h = \iota_X \circ (f \circ h) = \alpha_{Y',X}(h^*(f)).$$

This proves that the functor $T : \mathcal{C} \to \mathcal{D}$ is right adjoint to the inclusion functor $\mathcal{D} \hookrightarrow \mathcal{C}$.

The same proof gives a more general criterion that detects all adjunctions: Let $F : \mathcal{D} \to \mathcal{C}$ be any functor. Assume that for all $X \in \mathcal{C}$ there is some $G(X) \in \mathcal{D}$ equipped with a universal morphism $\varepsilon_X : F(G(X)) \to X$. The universal property says more precisely that for every morphism $F(Y) \to X$ there is a unique morphism $Y \to G(X)$ such that the evident triangle commutes. Then

• $G$ extends to a functor $G : \mathcal{C} \to \mathcal{D}$
• The bijections $\hom(Y,G(X)) \to \hom(F(Y),X)$, $f \mapsto \varepsilon_X \circ F(f)$ are natural,
• and hence, $G$ is right adjoint to $F$

By duality, there is a corresponding criterion for a left adjoint.

This criterion is very useful, because you will never need to verify naturality again when checking for an adjunction. I recommend this type of reasoning because you will spend less time with computations which are actually part of a more general phenomenon - and that is the whole purpose of category theory.

Answer 2

Your work and reasoning is very good so far. The only thing left to check is that we have naturality in $N$.

Say $g:N\to N',\,f:M\to t(N)$ are homomorphisms of Abelian groups. Let $j$ denote the natural inclusion $it\hookrightarrow\mathrm{id}_{\mathsf{Ab}}$.

We need to show that $j\circ t(g)\circ f=g\circ j\circ f:M\to N'$. Clearly it suffices to show $j\circ t(g)=g\circ j:t(N)\to N'$ i.e. show that $j$ is actually natural as I claimed above.

Well, $j\circ t(g)$ is the composition of the restriction and corestriction of $g$ to $t(N)$ and $t(N')$ with the inclusion $t(N')\hookrightarrow N'$; that's nothing but the restriction of $g$ to $t(N)$. On the other hand, $g\circ j$ is exactly the definition of the restriction of $g$ to $t(N)$, so we have equality.

As for intuition about bifunctors, bifunctors are just functors. Thankfully, thinking about bifunctors in terms of their slices/restrictions to single-variable functors is a safe thing to do. Really, the perceived complexity of bifunctors lies in the complexity of a product category, in diagrams of diagrams. I sometimes like to think of a bifunctor as a moving window, or as a machine that produces the same output as you move along a 'grid' (some arrows in the product category) in two different directions. That's probably not very helpful though.